Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e5 + 5;
const int N = 1e5;
int n, m, q, x, y, k, l[MAX], lg[MAX], p[MAX][20];
vector<int> g[MAX], v[MAX];
void dfs_lca(int u, int lev = 1, int pt = 0) {
  p[u][0] = pt;
  l[u] = lev;
  for (int v : g[u])
    if (v != pt) dfs_lca(v, lev + 1, u);
}
void init_lca() {
  dfs_lca(1);
  for (int i = 1; i <= n; ++i) lg[i] = log2(i);
  for (int j = 1; (1 << j) <= n; ++j)
    for (int i = 1; i <= n; ++i) p[i][j] = p[p[i][j - 1]][j - 1];
}
int lca(int a, int b) {
  if (l[a] < l[b]) swap(a, b);
  for (int i = lg[l[a]]; i >= 0; --i)
    if (l[a] - (1 << i) >= l[b]) a = p[a][i];
  if (a == b) return a;
  for (int i = lg[l[a]]; i >= 0; --i)
    if (p[a][i] && p[a][i] != p[b][i]) a = p[a][i], b = p[b][i];
  return p[a][0];
}
struct node {
  int v, l, r;
  node(int v, int l, int r) : v(v), l(l), r(r) {}
};
node null(0, 0, 0);
vector<node> St;
int root[MAX];
int build(int st, int nd) {
  if (st == nd) {
    St.push_back(null);
    return (int)St.size() - 1;
  }
  int piv = (st + nd) >> 1;
  int l = build(st, piv);
  int r = build(piv + 1, nd);
  St.push_back({0, l, r});
  return (int)St.size() - 1;
}
int update(int x, int st, int nd, int pos) {
  if (st > pos || nd < pos) return x;
  if (st == nd) {
    St.push_back({1, 0, 0});
    return (int)St.size() - 1;
  }
  int piv = (st + nd) >> 1;
  int l = update(St[x].l, st, piv, pos);
  int r = update(St[x].r, piv + 1, nd, pos);
  St.push_back({St[l].v + St[r].v, l, r});
  return (int)St.size() - 1;
}
void dfs(int nd, int pd = 0) {
  root[nd] = root[pd];
  for (int u : v[nd]) root[nd] = update(root[nd], 1, N, u);
  for (int u : g[nd])
    if (u != pd) dfs(u, nd);
}
int query(int x, int y, int z, int w, int st, int nd, int k, int lc) {
  int can = St[St[x].l].v + St[St[y].l].v - St[St[z].l].v - St[St[w].l].v;
  int can1 = St[x].v + St[y].v - St[z].v - St[w].v;
  int piv = (st + nd) >> 1;
  if (st == nd) {
    if (k == 1 && can1 == 1) return st;
    return -1;
  }
  if (can >= k)
    return query(St[x].l, St[y].l, St[z].l, St[w].l, st, piv, k, lc);
  return query(St[x].r, St[y].r, St[z].r, St[w].r, piv + 1, nd, k - can, lc);
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &x);
    v[x].push_back(i);
  }
  init_lca();
  St.push_back(null);
  root[0] = build(1, N);
  dfs(1, 0);
  while (q--) {
    scanf("%d %d %d", &x, &y, &k);
    int lc = lca(x, y);
    vector<int> sol;
    for (int i = 1; i <= k; ++i) {
      int s = query(root[x], root[y], root[lc], root[p[lc][0]], 1, N, i, lc);
      if (s == -1) break;
      sol.push_back(s);
    }
    printf("%d", (int)sol.size());
    for (int i = 0; i < (int)sol.size(); ++i) printf(" %d", sol[i]);
    printf("\n");
  }
  return 0;
}